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There is a coordinate-free description using only natural objects on $TM$. Here is one way to do it.

First, consider the basepoint submersion $\pi:TM\to M$. For each $v\in TM$, the linear map $\pi'(v):T_v(TM)\to T_{\pi(v)}M$ is surjective, and the $\pi$-fiber through $\pi(v)$ is equal to $T_{\pi(v)}M$, a vector space. It follows that the kernel of $\pi'(v)$ is naturally isomorphic to $T_{\pi(v)}M$. Call this isomorphism $\iota_v: T_{\pi(v)}M\to \mathrm{ker}\bigl(\pi'(v)\bigr)\subset T_v(TM)$.

Next, consider a Lagrangian $L:TM\to \mathbb{R}$, which I will assume to be smoothly differentiable. Define a $1$-form $\omega_L$ on $TM$ by $$ \omega_L(w) = dL\bigl( \iota_v(\pi'(v)(w))\bigr) $$ for all $w\in T_v(TM)$ and $v\in TM$. Also, let $R$ be the radial vector field on the fibers of $\pi:TM\to M$, and set $E_L = dL(R) - L$. (If $L$ is quadratic homogeneous, then, by Euler's relation, $E_L = L$.)

Finally, if $\gamma:[a,b]\to M$ is a twice differentiable curve, with $\gamma':[a,b]\to TM$ its velocity vector and $\gamma'':[a,b]\to T(TM)$ the velocity vector of $\gamma'$, then, for each $t\in[a,b]$ consider the co-vector $\beta(t)\in T^\ast_{\gamma'(t)}TM$ defined by the rule $$ \beta(t)(w) = d\omega_L(\gamma''(t),w)+ dE_L(w) $$ for $w\in T_{\gamma'(t)}TM$. It is not hard to show that $\beta(t)(w)=0$ for $w\in \mathrm{ker}\bigl(\pi'(\gamma'(t))\bigr)$, so it follows that $\beta(t) = \pi'(\gamma'(t))^\ast(\delta\gamma(t))$ for a unique co-vector $\delta\gamma(t)\in T^\ast_{\gamma(t)}M$.

This assignment $t\mapsto \delta\gamma(t)$ is the canonical 'variation $1$-form' of the Lagrangian $L$ along $\gamma$. It vanishes identically if and only if $\gamma$ satisfies the Euler-Lagrange equation for $L$.

There is a coordinate-free description using only natural objects on $TM$. Here is one way to do it.

First, consider the basepoint submersion $\pi:TM\to M$. For each $v\in TM$, the linear map $\pi'(v):T_v(TM)\to T_{\pi(v)}M$ is surjective, and the $\pi$-fiber through $v$ is equal to $T_{\pi(v)}M$, a vector space. It follows that the kernel of $\pi'(v)$ is naturally isomorphic to $T_{\pi(v)}M$. Call this isomorphism $\iota_v: T_{\pi(v)}M\to \mathrm{ker}\bigl(\pi'(v)\bigr)\subset T_v(TM)$, and let $\nu_v : T_v(TM) \to T_v(TM)$ be the nilpotent endomorphism $\nu_v = \iota_v\circ \pi'(v)$.

Next, consider a Lagrangian $L:TM\to \mathbb{R}$, which I will assume to be smoothly differentiable. Define a $1$-form $\omega_L$ on $TM$ by $$ \omega_L(w) = dL\bigl(\nu_v(w)\bigr) $$ for all $w\in T_v(TM)$. Let $R$ be the vector field on $TM$ that is tangent to the fibers of $\pi$ and that is the natural radial vector field on each such (vector space) fiber, and set $E_L = dL(R) - L$. (If $L$ is quadratic homogeneous on each $\pi$-fiber, then, by Euler's relation, $E_L = L$.)

Finally, if $\gamma:[a,b]\to M$ is a twice differentiable curve, with $\gamma':[a,b]\to TM$ its velocity vector and $\gamma'':[a,b]\to T(TM)$ the velocity vector of $\gamma'$, then, for each $t\in[a,b]$, consider the co-vector $\beta(t)\in T^\ast_{\gamma'(t)}TM$ defined by the rule $$ \beta(t)(w) = d\omega_L(\gamma''(t),w)+ dE_L(w) $$ for $w\in T_{\gamma'(t)}TM$. Then $\beta(t)(w)=0$ for $w\in \mathrm{ker}\bigl(\pi'(\gamma'(t))\bigr)$, so $\beta(t) = \pi'(\gamma'(t))^\ast(\delta\gamma(t))$ for a unique co-vector $\delta\gamma(t)\in T^\ast_{\gamma(t)}M$.

This assignment $t\mapsto \delta\gamma(t)$ is the canonical 'variation $1$-form' of the Lagrangian $L$ along $\gamma$. It vanishes identically if and only if $\gamma$ satisfies the Euler-Lagrange equation for $L$.

There is a coordinate-free description using only natural objects on $TM$. Here is one way to do it.

First, consider the basepoint submersion $\pi:TM\to M$. For each $v\in TM$, the linear map $\pi'(v):T_v(TM)\to T_{\pi(v)}M$ is surjective, and the $\pi$-fiber through $\pi(v)$ is equal to $T_{\pi(v)}M$, a vector space. It follows that the kernel of $\pi'(v)$ is naturally isomorphic to $T_{\pi(v)}M$. Call this isomorphism $\iota_v: T_{\pi(v)}M\to \mathrm{ker}\bigl(\pi'(v)\bigr)\subset T_v(TM)$.

Next, consider a Lagrangian $L:TM\to \mathbb{R}$, which I will assume to be smoothly differentiable. Define a $1$-form $\omega_L$ on $TM$ by $$ \omega_L(w) = dL\bigl( \iota_v(\pi'(v)(w))\bigr) $$ for all $w\in T_v(TM)$ and $v\in TM$. Also, let $R$ be the radial vector field on the fibers of $\pi:TM\to M$, and set $E_L = dL(R) - L$. (If $L$ is quadratic homogeneous, then, by Euler's relation, $E_L = L$.)

Finally, if $\gamma:[a,b]\to M$ is a twice differentiable curve, with $\gamma':[a,b]\to TM$ its velocity vector and $\gamma'':[a,b]\to T(TM)$ the velocity vector of $\gamma'$, then, for each $t\in[a,b]$ consider the co-vector $\beta(t)\in T^\ast_{\gamma'(t)}TM$ defined by the rule $$ \beta(t)(w) = d\omega_L(\gamma''(t),w)+ dE_L(w) $$ for $w\in T_{\gamma'(t)}TM$. It is not hard to show that $\beta(t)(w)=0$ for $w\in \mathrm{ker}\bigl(\pi'(\gamma'(t))\bigr)$, so it follows that $\beta(t) = \pi'(\gamma'(t))^\ast(\delta(\gamma(t)))$ for a unique co-vector $\delta(\gamma(t))\in T^\ast_{\gamma(t)}M$.

This assignment $t\mapsto \delta(\gamma(t))$ is the canonical 'variation $1$-form' of the Lagrangian $L$ along $\gamma$. It vanishes identically if and only if $\gamma$ satisfies the Euler-Lagrange equation for $L$.

There is a coordinate-free description using only natural objects on $TM$. Here is one way to do it.

First, consider the basepoint submersion $\pi:TM\to M$. For each $v\in TM$, the linear map $\pi'(v):T_v(TM)\to T_{\pi(v)}M$ is surjective, and the $\pi$-fiber through $\pi(v)$ is equal to $T_{\pi(v)}M$, a vector space. It follows that the kernel of $\pi'(v)$ is naturally isomorphic to $T_{\pi(v)}M$. Call this isomorphism $\iota_v: T_{\pi(v)}M\to \mathrm{ker}\bigl(\pi'(v)\bigr)\subset T_v(TM)$.

Next, consider a Lagrangian $L:TM\to \mathbb{R}$, which I will assume to be smoothly differentiable. Define a $1$-form $\omega_L$ on $TM$ by $$ \omega_L(w) = dL\bigl( \iota_v(\pi'(v)(w))\bigr) $$ for all $w\in T_v(TM)$ and $v\in TM$. Also, let $R$ be the radial vector field on the fibers of $\pi:TM\to M$, and set $E_L = dL(R) - L$. (If $L$ is quadratic homogeneous, then, by Euler's relation, $E_L = L$.)

Finally, if $\gamma:[a,b]\to M$ is a twice differentiable curve, with $\gamma':[a,b]\to TM$ its velocity vector and $\gamma'':[a,b]\to T(TM)$ the velocity vector of $\gamma'$, then, for each $t\in[a,b]$ consider the co-vector $\beta(t)\in T^\ast_{\gamma'(t)}TM$ defined by the rule $$ \beta(t)(w) = d\omega_L(\gamma''(t),w)+ dE_L(w) $$ for $w\in T_{\gamma'(t)}TM$. It is not hard to show that $\beta(t)(w)=0$ for $w\in \mathrm{ker}\bigl(\pi'(\gamma'(t))\bigr)$, so it follows that $\beta(t) = \pi'(\gamma'(t))^\ast(\delta\gamma(t))$ for a unique co-vector $\delta\gamma(t)\in T^\ast_{\gamma(t)}M$.

This assignment $t\mapsto \delta\gamma(t)$ is the canonical 'variation $1$-form' of the Lagrangian $L$ along $\gamma$. It vanishes identically if and only if $\gamma$ satisfies the Euler-Lagrange equation for $L$.

There is a coordinate-free description using just natural objects on $TM$. Here is one way to do it.

First, consider the base point submersion $\pi:TM\to M$. For each $v\in TM$, the map $\pi'(v):T_v(TM)\to T_{\pi(v)}M$ is a submersion, and the fiber through $\pi(v)$ is equal to $T_{\pi(v)}M$, a vector space. It follows that the kernel of $\pi'(v)$ is naturally isomorphic to $T_{\pi(v)}M$. Call this isomorphism $\iota_v: T_{\pi(v)}M\to \mathrm{ker}\bigl(\pi'(v)\bigr)$.

Now consider a Lagrangian $L:TM\to \mathbb{R}$. Define a $1$-form $\omega_L$ on $TM$ by $$ \omega_L(w) = dL\bigl( \iota_v(\pi'(v)(w))\bigr) $$ for all $w\in T_v(TM)$ and $v\in TM$. Also, let $R$ be the radial vector field on the fibers of $\pi:TM\to M$, and set $E_L = dL(R) - L$. (If $L$ is quadratic homogeneous, then, by Euler's relation, $E_L = L$.)

Finally, if $\gamma:[a,b]\to M$ is a twice differentiable curve, with $\gamma':[a,b]\to TM$ its velocity vector and $\gamma'':[a,b]\to T(TM)$ the velocity vector of $\gamma'$, then consider the $1$-form define by the rule $$ \beta(w) = d\omega_L(\gamma''(t),w)+ dE_L(w) $$ for $w\in T_{\gamma'(t)}TM$. It is not hard to show that $\beta(w)=0$ for $w\in \mathrm{ker}\bigl(\pi'(\gamma'(t))\bigr)$, so it follows that $\beta = (\pi'(\gamma'(t))^\ast(\delta(\gamma(t)))$ for a unique $\delta(\gamma(t))\in T^\ast_{\gamma(t)}M$.

This assignment $t\mapsto \delta(\gamma(t))$ is the canonical `variation' $1$-form. It vanishes identically if and only if $\gamma$ satisfies the Euler-Lagrange equation for $L$.

There is a coordinate-free description using only natural objects on $TM$. Here is one way to do it.

First, consider the basepoint submersion $\pi:TM\to M$. For each $v\in TM$, the linear map $\pi'(v):T_v(TM)\to T_{\pi(v)}M$ is surjective, and the $\pi$-fiber through $\pi(v)$ is equal to $T_{\pi(v)}M$, a vector space. It follows that the kernel of $\pi'(v)$ is naturally isomorphic to $T_{\pi(v)}M$. Call this isomorphism $\iota_v: T_{\pi(v)}M\to \mathrm{ker}\bigl(\pi'(v)\bigr)\subset T_v(TM)$.

Next, consider a Lagrangian $L:TM\to \mathbb{R}$, which I will assume to be smoothly differentiable. Define a $1$-form $\omega_L$ on $TM$ by $$ \omega_L(w) = dL\bigl( \iota_v(\pi'(v)(w))\bigr) $$ for all $w\in T_v(TM)$ and $v\in TM$. Also, let $R$ be the radial vector field on the fibers of $\pi:TM\to M$, and set $E_L = dL(R) - L$. (If $L$ is quadratic homogeneous, then, by Euler's relation, $E_L = L$.)

Finally, if $\gamma:[a,b]\to M$ is a twice differentiable curve, with $\gamma':[a,b]\to TM$ its velocity vector and $\gamma'':[a,b]\to T(TM)$ the velocity vector of $\gamma'$, then, for each $t\in[a,b]$ consider the co-vector $\beta(t)\in T^\ast_{\gamma'(t)}TM$ defined by the rule $$ \beta(t)(w) = d\omega_L(\gamma''(t),w)+ dE_L(w) $$ for $w\in T_{\gamma'(t)}TM$. It is not hard to show that $\beta(t)(w)=0$ for $w\in \mathrm{ker}\bigl(\pi'(\gamma'(t))\bigr)$, so it follows that $\beta(t) = \pi'(\gamma'(t))^\ast(\delta(\gamma(t)))$ for a unique co-vector $\delta(\gamma(t))\in T^\ast_{\gamma(t)}M$.

This assignment $t\mapsto \delta(\gamma(t))$ is the canonical 'variation $1$-form' of the Lagrangian $L$ along $\gamma$. It vanishes identically if and only if $\gamma$ satisfies the Euler-Lagrange equation for $L$.